List files and show them in a menu with bash

Question

I am trying to use the files in a directory as options in a bash script. The user should be able to select one and then pass the name of the selected file into a variable. So far I can get the list of files, but after a few hours trying I can't figure out how to show them as options.

#!/bin/bash
prompt="Please select a file:"
options=( $(find -maxdepth 1 -print0 | xargs -0) )

PS3="$prompt "
select opt in "${options[@]}" "Quit"; do 

    case "$REPLY" in
    for i in "${options[@]}"
    do
    $i' ) echo "You picked $opt which is file $REPLY";;'
    done    
    $(( ${#options[@]}+1 )) ) echo "Goodbye!"; break;;
    *) echo "Invalid option. Try another one.";continue;;

    esac

done

Any help is much appreciated. Thanks!

Answer 1

I do not think case is suitable here:

#!/bin/bash
prompt="Please select a file:"
options=( $(find -maxdepth 1 -print0 | xargs -0) )

PS3="$prompt "
select opt in "${options[@]}" "Quit" ; do 
    if (( REPLY == 1 + ${#options[@]} )) ; then
        exit

    elif (( REPLY > 0 && REPLY <= ${#options[@]} )) ; then
        echo  "You picked $opt which is file $REPLY"
        break

    else
        echo "Invalid option. Try another one."
    fi
done    

ls -ld $opt